integration - Showing the limit of $int_x^infty (y log y)^{-1}dy$ is zero

I'm trying to show

$$\lim_{x \to \infty} \int_x^\infty (y \log y)^{-1} dy = 0$$

In order to finish a proof. The problem I'm having is that without the limit, I know the integral diverges, and hence when I use substitution I end up with indeterminate form.

I think rather than using a substitution like $v = \log x$, I need to re-write the integral in the form of $e^{-t}$ so that the integral can be expressed in a proper form.

Any hints/advice is appreciated thank you!

Edit:

Using the substitution $t=y \log y$ I end up at an integral of the form $\int \frac{1}{1+e^t}dt$ which does not seem to help as I again end up with an indeterminate form

Answer

$\int \limits_{x}^{\infty} \frac{1}{y \ln y} dy > \int \limits_{x}^{x^2} \frac{1}{y \ln y} dy = \ln \ln x^2 -\ln \ln x = \ln 2 \approx 0.693 > 0$ for all $0 < x \in \mathbb{R}$, so if $\lim \limits_{x \to \infty} \int \limits_{x}^{\infty } \frac{1}{y \ln y} dy = 0$ then we would have that for all $x > x_0$ that $\int \limits_{ x}^{x^2} \frac{1}{y \ln y }d y < \epsilon$ which contradict the fact that $\int \limits_{x}^{x^2} \frac{1}{y \ln y} dy = \ln \ln x^2 -\ln \ln x = \ln 2 \approx 0.693$ and $\epsilon = 0.1$